Lecture 2: Enthalpy

1. Lecture 2: Enthalpy • Reading: Zumdahl 9.2, 9.3 • Outline • Definition of Enthalpy (H) • Definition of Heat Capacity (Cv and Cp) • Calculating E and H using Cv and Cp

2. What is Enthalpy? Thermodynamic Definition of Enthalpy (H) H = E + PV E = energy of the system P = pressure of the system V = volume of the system

3. Consider a process carried out at constant pressure: If work is of the form w= -PV, then: E = qp + w = qp - PV Rearranging, •  E + PV = qp where qp is heat transferred at constant pressure.

4. Recall that H = E + PV Then H = E + (PV) Since (PV) = PV + VP = PV (P constant) Substituting, H = E + PV But we just showed E + PV = qp So finally, H = qp In words, the change in enthalpy (H) is equal to the heat transferred at constant pressure (qp)

5. What about changes in enthalpy? Consider the following expression for enthalpy change in a chemical process: H = Hprods - Hreacts If H >0, then qp >0, and the reaction is endothermic. If H <0, then qp <0, and the reaction is exothermic.

6. Picturing Enthalpy Changes (Similar to previous discussion for energy changes) If heat comes out of system, the enthalpy decreases (ex. cooling water). If heat goes into the system, the enthalpy increases (ex. heating water)

7. Heat capacity at constant volume • Recall from Chapter 5 (section 5.6): (KE)ave = (3/2) RT (for an ideal monatomic gas) • Temperature is a measure of molecular speed. • In thermodynamic terms, an increase in system temperature corresponds to an increase in average kinetic energy of a gas ( i.e., T is proportional to KE)

8. Heat capacity at constant volume • (KE)ave = 3/2 RT (ideal monatomic gas) • How much energy in the form of heat is required to change the gas temperature by an amount T? Heat required = 3/2RT = 3/2R (for T = 1K) • Therefore, Cv = 3/2 R is the heat required to raise one mole of an ideal gas by 1K at constant volume. Cv is referred to as the constant volume heat capacity.

9. Q: Heat capacity at constant P? What about at constant pressure? In this case, (PV)-type work can also occur: PV = nRT PV = nRT = RT (for 1 mole) = R (for T = 1 K) Cp = “heat into translation” + “work” = Cv + R = 5/2R (for an ideal monatomic gas)

10. Cv for Monatomic Gases Q: What are the energetic degrees of freedom for a monatomic gas? Ans: Just translations(x,y,z), which contribute (3/2)R to Cv.

11. Cv for Polyatomics Q: What are the energetic degrees of freedom for a polyatomic gas? Ans: translations, and rotations, and vibrations… all of which may contribute to Cv (depends on T). 3

12. Variation in Cp and Cv • Monatomics: Cv = 3/2 R Cp = 5/2 R • Polyatomics: Cv > 3/2 R Cp > 5/2 R But, Cp = Cv + R (always!)

13. Heating an ideal monatomic gas at constant volume Recall Eave = 3/2 nRT (av trans. energy) So E = 3/2 nRT = (3/2R)nT Recall Cv = 3/2 R So, E = nCvT Q: Why Cv? We envision heating our system at constant volume. As such, all heat goes towards increasing E (no work is done, since no volume change is possible)

14. What if we heat the ideal gas at constant pressure? In this case, we have a volume change and (PV) work can occur: recall Cp = Cv + R q p = n CpT q p = n (Cv + R) T recall E = nCvT so q p = E + nRT = E + PV q p = H = n CpT

15. Ideal Monatomic Gas Cv = 3/2R Cp = Cv + R = 5/2 R Polyatomic Gas Cv > 3/2R Cp > 5/2 R All Ideal Gases E = nCvT H = nCpT Keeping Track

16. Example What is q, w, E and H for a process in which one mole of an ideal monatomic gas with an initial volume of 5 liters and pressure of 2.0 atm is heated until a volume of 10 liters is reached with pressure unchanged? Pinit = 2 atm Pfinal = 2 atm Vinit = 5 l Vfinal = 10 l Tinit = ? K Tfinal = ? K

17. Using the ideal gas law PV = nRT PV = nRT, so we can determine T. V = (10 L - 5 L) = 5 L Substituting values,

18. So, now we can calculate ∆E, ∆H, w and q: